Topic 4 Notes
Jeremy Orloff
4
4.1
Cauchy’s integral formula
Introduction
Cauchy’s theorem is a big theorem which we will use almost daily from here on out. Right
away it will reveal a number of interesting and useful properties of analytic functions. More
will follow as the course progresses.
If you learn just one theorem this week it should be Cauchy’s integral formula!
We start with a statement of the theorem for functions. After some examples, we’ll give a
generalization to all derivatives of a function. After some more examples we will prove the
theorems. After that we will see some remarkable consequences that follow fairly directly
from the Cauchy’s formula.
4.2
Cauchy’s integral for functions
Theorem 4.1. (Cauchy’s integral formula) Suppose C is a simple closed curve and the
function f (z) is analytic on a region containing C and its interior. We assume C is oriented
counterclockwise. Then for any z0 inside C:
1
f (z0 ) =
2πi
Z
C
f (z)
dz
z − z0
(1)
Im(z)
A
C
z0
Re(z)
Cauchy’s integral formula: simple closed curve C, f (z) analytic on and inside C.
This is remarkable: it says that knowing the values of f on the boundary curve C means we
know everything about f inside C!! This is probably unlike anything you’ve encountered
with functions of real variables.
Aside 1. With a slight change of notation (z becomes w and z0 becomes z) we often write
the formula as
Z
f (w)
1
f (z) =
dw
(2)
2πi C w − z
1
4
2
CAUCHY’S INTEGRAL FORMULA
Aside 2. We’re not being entirely fair to functions of real variables. We will see that for
f = u + iv the real and imaginary parts u and v have many similar remarkable properties.
u and v are called conjugate harmonic functions.
4.2.1
Examples
Z
Example 4.2. Compute
c
2
ez
dz, where C is the curve shown.
z−2
Im(z)
C
Re(z)
2
2
Solution: Let f (z) = ez . f (z) is entire. Since C is a simple closed curve (counterclockwise)
and z = 2 is inside C, Cauchy’s integral formula says that the integral is 2πif (2) = 2πie4 .
Example 4.3. Do the same integral as the previous example with C the curve shown.
Im(z)
C
Re(z)
2
2
Solution: Since f (z) = ez /(z − 2) is analytic on and inside C, Cauchy’s theorem says that
the integral is 0.
Example 4.4. Do the same integral as the previous examples with C the curve shown.
Im(z)
C
Re(z)
2
2
Solution: This one is trickier. Let f (z) = ez . The curve C goes around 2 twice in the
clockwise direction, so we break C into C1 + C2 as shown in the next figure.
4
3
CAUCHY’S INTEGRAL FORMULA
Im(z)
C1
C1
2
Re(z)
C2
These are both simple closed curves, so we can apply the Cauchy integral formula to each
separately. (The negative signs are because they go clockwise around z = 2.)
Z
Z
Z
f (z)
f (z)
f (z)
dz =
dz +
dz = −2πif (2) − 2πif (2) = −4πif (2).
C z−2
C1 z − 2
C2 z − 2
4.3
Cauchy’s integral formula for derivatives
Cauchy’s integral formula is worth repeating several times. So, now we give it for all
derivatives f (n) (z) of f . This will include the formula for functions as a special case.
Theorem 4.5. Cauchy’s integral formula for derivatives. If f (z) and C satisfy the same
hypotheses as for Cauchy’s integral formula then, for all z inside C we have
Z
n!
f (w)
(n)
f (z) =
dw, n = 0, 1, 2, . . .
(3)
2πi C (w − z)n+1
where, C is a simple closed curve, oriented counterclockwise, z is inside C and f (w) is
analytic on and inside C.
Z 2z
e
Example 4.6. Evaluate I =
dz where C : |z| = 1.
4
C z
Solution: With Cauchy’s formula for derivatives this is easy. Let f (z) = e2z . Then,
Z
f (z)
2πi 000
8
I=
dz =
f (0) = πi.
4
3!
3
C z
Example 4.7. Now Let C be the contour shown below and evaluate the same integral as
in the previous example.
Im
C
Re
8
Solution: Again this is easy: the integral is the same as the previous example, i.e. I = πi.
3
4
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CAUCHY’S INTEGRAL FORMULA
4.3.1
Another approach to some basic examples
Suppose C is a simple closed curve around 0. We have seen that
Z
1
dz = 2πi.
C z
The Cauchy integral formula gives the same result. That is, let f (z) = 1, then the formula
says
Z
1
f (z)
dz = f (0) = 1.
2πi C z − 0
Likewise Cauchy’s formula for derivatives shows
Z
Z
1
f (z)
dz
=
dz = f (n) (0) = 0,
n
n+1
(z)
z
C
C
4.3.2
for integers n > 1.
More examples
Z
Example 4.8. Compute
C
cos(z)
dz
z(z 2 + 8)
over the contour shown.
Im(z)
2i
C
Im(z)
−2i
Solution: Let f (z) = cos(z)/(z 2 + 8). f (z) is analytic on and inside the curve C. That is,
the roots of z 2 + 8 are outside the curve. So, we rewrite the integral as
Z
Z
cos(z)/(z 2 + 8)
f (z)
1
πi
dz =
dz = 2πif (0) = 2πi = .
z
8
4
C
C z
Z
Example 4.9. Compute
C
1
dz
(z 2 + 4)2
over the contour shown.
4
5
CAUCHY’S INTEGRAL FORMULA
Im(z)
2i
C
i
Re(z)
−i
−2i
Solution: We factor the denominator as
(z 2
1
1
=
.
2
2
+ 4)
(z − 2i) (z + 2i)2
Let
f (z) =
1
(z + 2i)2
. Clearly f (z) is analytic inside C. So, by Cauchy’s formula for derivatives:
Z
Z
1
f (z)
−2
4πi
π
0
dz =
= 2πif (2i) = 2πi
=
=
2 + 4)2
2
3
(z
(z
−
2i)
(z
+
2i)
64i
16
C
C
z=2i
Z
Example 4.10. Compute
C
z2
z
dz
+4
over the curve C shown below.
Im(z)
2i
i
C
Re(z)
−i
−2i
Solution: The integrand has singularities at ±2i and the curve C encloses them both. The
solution to the previous solution won’t work because we can’t find an appropriate f (z) that
is analytic on the whole interior of C. Our solution is to split the curve into two pieces.
Notice that C3 is traversed both forward and backward.
4
6
CAUCHY’S INTEGRAL FORMULA
Im(z)
2i
C1
C1
i
C3
Re(z)
−C3
−i
C2
C2
−2i
Split the original curve C into 2 pieces that each surround just one singularity.
We have
We let
z
z
=
.
z2 + 4
(z − 2i)(z + 2i)
f1 (z) =
So,
z
z + 2i
and f2 (z) =
z
.
z − 2i
z
f1 (z)
f2 (z)
=
=
.
+4
z − 2i
z + 2i
The integral, can be written out as
Z
Z
Z
Z
z
f1 (z)
f2 (z)
z
dz =
dz =
dz +
dz
2
2
C1 +C3 −C3 +C2 z + 4
C1 +C3 z − 2i
C2 −C3 z + 2i
C z +4
z2
Since f1 is analytic inside the simple closed curve C1 + C3 and f2 is analytic inside the
simple closed curve C2 − C3 , Cauchy’s formula applies to both integrals. The total integral
equals
2πi(f1 (2i) + f2 (−2i)) = 2πi(1/2 + 1/2) = 2πi.
Remarks. 1. We could also have done this problem using partial fractions:
z
A
B
=
+
.
(z − 2i)(z + 2i)
z − 2i z + 2i
It will turn out that A = f1 (2i) and B = f2 (−2i). It is easy to apply the Cauchy integral
formula to both terms.
2. Important note. In an upcoming topic we will formulate the Cauchy residue theorem.
This will allow us to compute the integrals in Examples 4.8-4.10 in an easier and less ad
hoc manner.
4
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CAUCHY’S INTEGRAL FORMULA
4.3.3
The triangle inequality for integrals
We discussed the triangle inequality in the Topic 1 notes. It says that
|z1 + z2 | ≤ |z1 | + |z2 |,
(4a)
with equality if and only if z1 and z2 lie on the same ray from the origin.
A useful variant of this statement is
|z1 | − |z2 | ≤ |z1 − z2 |.
(4b)
This follows because Equation 4a implies
|z1 | = |(z1 − z2 ) + z2 | ≤ |z1 − z2 | + |z2 |.
Now subtracting z2 from both sides give Equation 4b
Since an integral is basically a sum, this translates to the triangle inequality for integrals.
We’ll state it in two ways that will be useful to us.
Theorem 4.11. (Triangle inequality for integrals) Suppose g(t) is a complex valued function of a real variable, defined on a ≤ t ≤ b. Then
Z
b
a
g(t) dt ≤
b
Z
a
|g(t))| dt,
with equality if and only if the values of g(t) all lie on the same ray from the origin.
Proof. This follows by approximating the integral as a Riemann sum.
Z
a
b
g(t) dt ≈
X
g(tk )∆t ≤
X
|g(tk )|∆t ≈
Z
a
b
|g(t)| dt.
The middle inequality is just the standard triangle inequality for sums of complex numbers. Theorem 4.12. (Triangle inequality for integrals II) For any function f (z) and any curve
γ, we have
Z
Z
f (z) dz ≤ |f (z)| |dz|.
γ
γ
Here dz =
γ 0 (t) dt
and |dz| =
|γ 0 (t)| dt.
Proof. This follows immediately from the previous theorem:
Z
Z
f (z) dz =
γ
a
b
0
f (γ(t))γ (t) dt ≤
Z
a
b
0
|f (γ(t))||γ (t)| dt =
Corollary. If |f (z)| < M on C then
Z
f (z) dz ≤ M · (length of C).
C
Z
γ
|f (z)| |dz|.
4
8
CAUCHY’S INTEGRAL FORMULA
Proof. Let γ(t), with a ≤ t ≤ b, be a parametrization of C. Using the triangle inequality
Z
C
f (z) dz ≤
Z
C
b
Z
|f (z)| |dz| =
a
|f (γ(t))| |γ 0 (t)| dt ≤
b
Z
a
M |γ 0 (t)| dt = M · (length of C).
Here we have used that
|γ 0 (t)| dt =
p
(x0 )2 + (y 0 )2 dt = ds,
the arclength element. Example 4.13. Compute the real integral
Z ∞
I=
(x2
−∞
1
dx
+ 1)2
Solution: The trick is to integrate f (z) = 1/(z 2 + 1)2 over the closed contour C1 + CR
shown, and then show that the contribution of CR to this integral vanishes as R goes to ∞.
Im(z)
CR
CR
i
−R
C1
The only singularity of
f (z) =
R
Re(z)
1
(z +
i)2 (z
inside the contour is at z = i. Let
g(z) =
− i)2
1
.
(z + i)2
Since g is analytic on and inside the contour, Cauchy’s formula gives
Z
Z
−2
π
g(z)
f (z) dz =
dz = 2πig 0 (i) = 2πi
= .
2
3
(2i)
2
C1 +CR
C1 +CR (z − i)
We parametrize C1 by
γ(x) = x,
with
− R ≤ x ≤ R.
So,
Z
Z
R
f (z) dz =
C1
−R
(x2
1
dx.
+ 1)2
This goes to I (the value we want to compute) as R → ∞.
Next, we parametrize CR by
γ(θ) = Reiθ ,
with
0 ≤ θ ≤ π.
4
9
CAUCHY’S INTEGRAL FORMULA
So,
Z
π
Z
f (z) dz =
CR
0
1
(R2 e2iθ
+ 1)2
iReiθ dθ
By the triangle inequality for integrals, if R > 1
Z
Z π
1
iReiθ dθ.
f (z) dz ≤
2
2iθ
(R e + 1)2
CR
0
(5)
From the triangle equality in the form Equation 4b we know that
|R2 e2iθ + 1| ≥ |R2 e2iθ | − |1| = R2 − 1.
Thus,
1
1
≤ 2
R −1
|R2 e2iθ + 1|
⇒
1
1
≤
.
(R2 − 1)2
|R2 e2iθ + 1|2
Using Equation 5, we then have
Z
Z π
Z π
R
πR
1
iθ
f (z) dz ≤
iRe dθ ≤
dθ =
2
2
2
2
2iθ
2
(R − 1)2
(R e + 1)
CR
0 (R − 1)
0
Clearly this goes to 0 as R goes to infinity. Thus, the integral over the contour C1 + CR
goes to I as R gets large. But
Z
f (z) dz = π/2
C1 +CR
for all R > 1. We can therefore conclude that I = π/2.
As a sanity check, we note that our answer is real and positive as it needs to be.
4.4
4.4.1
Proof of Cauchy’s integral formula
A useful theorem
Before proving the theorem we’ll need a theorem that will be useful in its own right.
Theorem 4.14. (A second extension of Cauchy’s theorem) Suppose that A is a simply
connected region containing the point z0 . Suppose g is a function which is
1. Analytic on A − {z0 }
2. Continuous on A. (In particular, g does not blow up at z0 .)
Then,
Z
g(z) dz = 0
C
for all closed curves C in A.
Proof. The extended version of Cauchy’s theorem in the Topic 3 notes tells us that
Z
Z
g(z) dz =
g(z) dz,
C
Cr
4
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CAUCHY’S INTEGRAL FORMULA
where Cr is a circle of radius r around z0 .
Im(z)
A
z0
Cr
C
Re(z)
Since g(z) is continuous we know that |g(z)| is bounded inside Cr . Say, |g(z)| < M . The
corollary to the triangle inequality says that
Z
g(z) dz ≤ M (length of Cr ) = M 2πr.
Cr
Since r can be as small as we want, this implies that
Z
g(z) dz = 0.
Cr
Note. Using this, we can show that g(z) is, in fact, analytic at z0 . The proof will be the
same as in our proof of Cauchy’s theorem that g(z) has an antiderivative.
4.4.2
Proof of Cauchy’s integral formula
We reiterate Cauchy’s integral formula from Equation 1:
1
f (z0 ) =
2πi
Z
C
f (z)
dz.
z − z0
Im(z)
A
C
z0
Re(z)
Proof. (of Cauchy’s integral formula) We use a trick that is useful enough to be worth
remembering. Let
f (z) − f (z0 )
g(z) =
.
z − z0
Since f (z) is analytic on A, we know that g(z) is analytic on A − {z0 }. Since the derivative
of f exists at z0 , we know that
lim g(z) = f 0 (z0 ).
z→z0
4
11
CAUCHY’S INTEGRAL FORMULA
That is, if we define g(z0 ) = f 0 (z0 ) then g is continuous at z0 . From the extension of
Cauchy’s theorem just above, we have
Z
Z
f (z) − f (z0 )
g(z) dz = 0, i.e.
dz = 0.
z − z0
C
C
Thus
Z
C
f (z)
dz =
z − z0
Z
C
f (z0 )
dz = f (z0 )
z − z0
Z
C
1
dz = 2πif (z0 ).
z − z0
The last equality follows from our, by now, well known integral of 1/(z − z0 ) on a loop
around z0 .
4.5
Proof of Cauchy’s integral formula for derivatives
Recall that Cauchy’s integral formula in Equation 3 says
Z
n!
f (w)
(n)
f (z) =
dw, n = 0, 1, 2, . . .
2πi C (w − z)n+1
First we’ll offer a quick proof which captures the reason behind the formula, and then a
formal proof.
Quick proof: We have an integral representation for f (z), z ∈ A, we use that to find an
integral representation for f 0 (z), z ∈ A.
Z
Z
Z
d
1
f (w)
1
d
f (w)
1
f (w)
0
f (z) =
dw =
dw =
dw
dz 2πi C w − z
2πi C dz w − z
2πi C (w − z)2
(Note, since z ∈ A and w ∈ C, we know that w − z 6= 0) Thus,
Z
1
f (w)
0
f (z) =
dw
2πi C (w − z)2
Now, by iterating this process, i.e. by mathematical induction, we can show the formula
for higher order derivatives.
Formal proof: We do this by taking the limit of
lim
∆z→0
f (z + ∆z) − f (z)
∆z
using the integral representation of both terms:
Z
1
f (w)
f (z + ∆z) =
dw,
2πi C w − z − ∆z
1
f (z) =
2πi
Z
C
f (w)
dw
w−z
Now, using a little algebraic manipulation we get
Z
f (z + ∆z) − f (z)
1
f (w)
f (w)
=
−
dw
∆z
2πi ∆z C w − z − ∆z w − z
Z
1
f (w)∆z
=
dw
2πi ∆z C (w − z − ∆z)(w − z)
Z
1
f (w)
=
dw
2πi C (w − z)2 − ∆z(w − z)
4
CAUCHY’S INTEGRAL FORMULA
12
Letting ∆z go to 0, we get Cauchy’s formula for f 0 (z):
Z
1
f (w)
0
f (z) =
dw
2πi C (w − z)2
There is no problem taking the limit under the integral sign because everything is continuous
and the denominator is never 0. 4.6
4.6.1
Amazing consequence of Cauchy’s integral formula
Existence of derivatives
Theorem. Suppose f (z) is analytic on a region A. Then, f has derivatives of all order.
Proof. This follows from Cauchy’s integral formula for derivatives. That is, we have a
formula for all the derivatives, so in particular the derivatives all exist.
A little more precisely: for any point z in A we can put a small disk around z0 that is
entirely contained in A. Let C be the boundary of the disk, then Cauchy’s formula gives
a formula for all the derivatives f (n) (z0 ) in terms of integrals over C. In particular, those
derivatives exist. Remark. If you look at the proof of Cauchy’s formula for derivatives you’ll see that f
having derivatives of all orders boils down to 1/(w − z) having derivatives of all orders for
w on a curve not containing z.
Important remark. We have at times assumed that for f = u + iv analytic, u and v have
continuous higher order partial derivatives. This theorem confirms that fact. In particular,
uxy = uyx , etc.
4.6.2
Cauchy’s inequality
Theorem 4.15. (Cauchy’s inequality) Let CR be the circle |z − z0 | = R. Assume that f (z)
is analytic on CR and its interior, i.e. on the disk |z − z0 | ≤ R. Finally let MR = max |f (z)|
over z on CR . Then
n!MR
|f (n) (z0 )| ≤
, n = 1, 2, 3, . . .
(6)
Rn
Proof. Using Cauchy’s integral formula for derivatives (Equation 3) we have
Z
Z
n!
|f (w)|
n! MR
n! MR
(n)
|f (z0 )| ≤
|dw| ≤
|dw| =
· 2πR
n+1
n+1
2π CR |w − z0 |
2π R
2π Rn+1
CR
4.6.3
Liouville’s theorem
Theorem 4.16. (Liouville’s theorem) Assume f (z) is entire and suppose it is bounded in
the complex plane, namely |f (z)| < M for all z ∈ C then f (z) is constant.
M
Proof. For any circle of radius R around z0 the Cauchy inequality says |f 0 (z0 )| ≤
. But,
R
R can be as large as we like so we conclude that |f 0 (z0 )| = 0 for every z0 ∈ C. Since the
derivative is 0, the function itself is constant.
4
13
CAUCHY’S INTEGRAL FORMULA
In short:
If f is entire and bounded then f is constant.
Note. P (z) = an z n + . . . + a0 , sin(z), ez are all entire but not bounded.
Now, practically for free, we get the fundamental theorem of algebra.
Corollary. (Fundamental theorem of algebra) Any polynomial P of degree n ≥ 1, i.e.
P (z) = a0 + a1 z + . . . + an z n , an 6= 0,
has exactly n roots.
Proof. There are two parts to the proof.
Hard part: Show that P has at least one root.
This is done by contradiction, together with Liouville’s theorem. Suppose P (z) does not
have a zero. Then
1. f (z) = 1/P (z) is entire. This is obvious because (by assumption) P (z) has no zeros.
2. f (z) is bounded. This follows because 1/P (z) goes to 0 as |z| goes to ∞.
Im(z)
|1/P (z)| small out here.
CR
R
Im(z)
M = max of |1/P (z)| in here.
(It is clear that |1/P (z)| goes to 0 as z goes to infinity, i.e. |1/P (z)| is small outside a large
circle. So |1/P (z)| is bounded by M .)
So, by Liouville’s theorem f (z) is constant, and therefore P (z) must be constant as well.
But this is a contradiction, so the hypothesis of “No zeros” must be wrong, i.e. P must
have a zero.
Easy part: P has exactly n zeros. Let z0 be one zero. We can factor P (z) = (z − z0 )Q(z).
Q(z) has degree n − 1. If n − 1 > 0, then we can apply the result to Q(z). We can continue
this process until the degree of Q is 0.
4.6.4
Maximum modulus principle
Briefly, the maximum modulus principle states that if f is analytic and not constant in
a domain A then |f (z)| has no relative maximum in A and the absolute maximum of |f |
occurs on the boundary of A.
4
14
CAUCHY’S INTEGRAL FORMULA
In order to prove the maximum modulus principle we will first prove the mean value property. This will give you a good feel for the maximum modulus principle. It is also important
and interesting in its own right.
Theorem 4.17. (Mean value property) Suppose f (z) is analytic on the closed disk of
radius r centered at z0 , i.e. the set |z − z0 | ≤ r. Then,
1
f (z0 ) =
2π
2π
Z
f (z0 + reiθ ) dθ
(7)
0
Proof. This is an application of Cauchy’s integral formula on the disk Dr = |z − z0 | ≤ r.
Im(z)
r
Cr
z0
Re(z)
We can parametrize Cr , the boundary of Dr , as
γ(t) = z0 + reiθ , with 0 ≤ θ ≤ 2π, so γ 0 (θ) = ireiθ .
By Cauchy’s formula we have
Z
Z 2π
Z 2π
1
f (z)
1
f (z0 + reiθ )
1
iθ
f (z0 ) =
dz =
ire dθ =
f (z0 + reiθ ) dθ
2πi Cr z − z0
2πi 0
2π 0
reiθ
This proves the property. In words, the mean value property says f (z0 ) is the arithmetic mean of the values on the
circle.
Now we can state and prove the maximum modulus principle. We state the assumptions
carefully. When applying this theorem, it is important to verify that the assumptions are
satisfied.
Theorem 4.18. (Maximum modulus principle) Suppose f (z) is analytic in a connected
region A and z0 is a point in A.
1. If |f | has a relative maximum at z0 then f (z) is constant in a neighborhood of z0 .
2. If A is bounded and connected, and f is continuous on A and its boundary, then either
f is constant or the absolute maximum of |f | occurs only on the boundary of A.
Proof. Part (1): The argument for part (1) is a little fussy. We will use the mean value
property and the triangle inequality from Theorem 4.11.
Since z0 is a relative maximum of |f |, for every small enough circle C : |z − z0 | = r around
z0 we have |f (z)| ≤ |f (z0 )| for z on C. Therefore, by the mean value property and the
4
15
CAUCHY’S INTEGRAL FORMULA
triangle inequality
Z 2π
1
f (z0 + reiθ ) dθ
|f (z0 )| =
2π 0
Z 2π
1
≤
|f (z0 + reiθ )| dθ
2π 0
Z 2π
1
≤
|f (z0 )| dθ
2π 0
= |f (z0 )|
(mean value property)
(triangle inequality)
(|f (z0 + reiθ )| ≤ |f (z0 )|)
Since the beginning and end of the above are both |f (z0 )| all the inequalities in the chain
must be equalities.
The first inequality can only be an equality if for all θ, f (z0 + reiθ ) lie on the same ray from
the origin, i.e. have the same argument or are 0.
The second inequality can only be an equality if all |f (z0 + reiθ )| = |f (z0 )|. So we have all
f (z0 + reiθ ) have the same magnitude and the same argumeny. This implies they are all
the same.
Finally, if f (z) is constant along the circle and f (z0 ) is the average of f (z) over the circle
then f (z) = f (z0 ), i.e. f is constant on a small disk around z0 .
Part (2): The assumptions that A is bounded and f is continuous on A and its boundary
serve to guarantee that |f | has an absolute maximum (on A combined with its boundary).
Part (1) guarantees that the absolute maximum can not lie in the interior of the region A
unless f is constant. (This requires a bit more argument. Do you see why?) If the absolute
maximum is not in the interior it must be on the boundary. Example 4.19. Find the maximum modulus of ez on the unit square with 0 ≤ x, y ≤ 1.
Solution:
|ex+iy | = ex ,
so the maximum is when x = 1, 0 ≤ y ≤ 1 is arbitrary. This is indeed on the boundary of
the unit square
Example 4.20. Find the maximum modulus for sin(z) on the square [0, 2π] × [0, 2π].
Solution: We use the formula
sin(z) = sin x cosh y + i cos x sinh y.
So,
| sin(z)|2 = sin2 x cosh2 y + cos2 x sinh2 y
= sin2 x cosh2 y + (1 − sin2 x) sinh2 y
= sin2 x + sinh2 y.
We know the maximum over x of sin2 (x) is at x = π/2 and x = 3π/2. The maximum of
sinh2 y is at y = 2π. So maximum modulus is
q
q
2
1 + sinh (2π) = cosh2 (2π) = cosh(2π).
4
16
CAUCHY’S INTEGRAL FORMULA
This occurs at the points
z = x + iy =
π
+ 2πi,
2
and z =
3π
+ 2πi.
2
Both these points are on the boundary of the region.
Example 4.21. Suppose f (z) is entire. Show that if lim f (z) = 0 then f (z) ≡ 0.
z→∞
Solution: This is a standard use of the maximum modulus principle. The strategy is to show
that the maximum of |f (z)| is not on the boundary (of the appropriately chosen region), so
f (z) must be constant.
Fix z0 . For R > |z0 | let MR be the maximum of |f (z)| on the circle |z| = R. The maximum
modulus theorem says that |f (z0 )| < MR . Since f (z) goes to 0, as R goes to infinity, we
must have MR also goes to 0. This means |f (z0 )| = 0. Since this is true for any z0 , we have
f (z) ≡ 0.
Example 4.22. Here is an example of why you need A to be bounded in the maximum
modulus theorem. Let A be the upper half-plane
Im(z) > 0.
So the boundary of A is the real axis.
Let f (z) = e−iz . We have
|f (x)| = |e−ix | = 1
for x along the real axis. Since |f (2i)| = |e2 | > 1, we see |f | cannot take its maximum along
the boundary of A.
Of course, it can’t take its maximum in the interior of A either. What happens here is that
f (z) doesn’t have a maximum modulus. Indeed |f (z)| goes to infinity along the positive
imaginary axis.